My work so far: Since the lines are perpendicular, the dot product of their direction vectors should be $0$, so $<1,-1,2>\cdot <x,y,z>=0$.
But I'm not sure where to go from here. I don't know how to express the fact that the line goes through $(0,1,2)$, or that it intersects the other line. How should I approach this problem?
Best Answer
Hint. Suppose that the required line intersects the given line at the point with parameter $a$, that is, $$(1+a,\,1-a,\,2a)\ .$$ Then the direction of the required line is $$(1+a,\,1-a,\,2a)-(0,1,2)=(1+a,\,-a,\,2a-2)\ .$$ This must be perpendicular to the direction of the given line, which is $(1,-1,2)$. Therefore $$(1+a,\,-a,\,2a-2)\cdot(1,-1,2)=0\ .$$ Can you use this equation to find $a$ and then finish the problem?
Good luck!
Alternative method. Let ${\bf u}=(1,-1,2)$ be the direction of the given line. A vector in the plane of both lines is $${\bf v}=(1,1,0)-(0,1,2)=(1,0,-2)\ .$$ A vector perpendicular to the plane containing both lines is $${\bf n}={\bf u}\times{\bf v}$$ and the required line is perpendicular to both $\bf u$ and $\bf n$, so its direction is $${\bf u}\times{\bf n}={\bf u}\times({\bf u}\times{\bf v})\ .$$